[Big-Daddy]

In the reaction, P + Q  R + S, the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure.

(A straight-line graph of concentration of Q (y-axis) against time (x-axis) with positive y-intercept Q₀.)

What is the overall order of the reaction?

I know the question-writer wants me to start working out the order of P and then take the order of Q to be 0. But if P and Q are both reactants, and the graph of Q against time is a straight-line, surely this means the overall order is 0 - so must be 0 for both P and Q? Or does the question, in its unclear way, imply that P has to be in excess?

[mjc123]

This makes no sense, at face value. Whatever the rate law, by stoichiometry dP/dt = dQ/dt, which isn't the case, as P seems to be decreasing exponentially and Q linearly. Are you sure it isn't describing two experiments - one measuring P vs t with Q in excess, the other measuring Q vs t with P in excess? That would be consistent with an order of 1 in P and 0 in Q.

[Big-Daddy]

This makes no sense, at face value. Whatever the rate law, by stoichiometry dP/dt = dQ/dt, which isn't the case, as P seems to be decreasing exponentially and Q linearly. Are you sure it isn't describing two experiments - one measuring P vs t with Q in excess, the other measuring Q vs t with P in excess? That would be consistent with an order of 1 in P and 0 in Q.

Thanks. There is no hint that it's describing such a case but I guess it's inevitable, since there's no way Q could be linearly dependent if P is exponentially dependent in the same time-frame.

Something interesting I found - I tried calculating n using the data given in the first line of the first post about P, but taking the nth-order integrated rate law for n≠1 so we'd expect it to crash at some point. I end up with 0.25^1-n - 1 = 2*(0.5^1-n - 1). I can't see an obvious way to solve this but it seems clear from inspection that the solution - the only solution - is n=1. Nice to see n can be found this way for any n, even n=1, despite the integration method precluding n=1.

[mjc123]

Unfortunately this is a trivial solution, as both sides are identically zero, whatever the numbers you plug in, since x⁰ = 1 for all x. Even for, say a zero-order reaction where t(75%) = 1.5*t(50%), you will find n=1 is a trivial solution, as well as the non-trivial n=0. (By the way, I think the exponents in your equation should be -(n-1).)

[Big-Daddy]

Unfortunately this is a trivial solution, as both sides are identically zero, whatever the numbers you plug in, since x⁰ = 1 for all x. Even for, say a zero-order reaction where t(75%) = 1.5*t(50%), you will find n=1 is a trivial solution, as well as the non-trivial n=0. (By the way, I think the exponents in your equation should be -(n-1).)

Ah of course.  I should have noticed it doesn't matter what fractions are used or what the ratio of times is, so n=1 must be a trivial solution.

And -(n-1)=1-n, is that what you mean?